Three circles of radius 2 are placed such that each of them touches the other. The area of the region enclosed by the three circles has to be determined. The circles are tangent to each other.

It is seen that if the radius of any of circles touching the point...

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Three circles of radius 2 are placed such that each of them touches the other. The area of the region enclosed by the three circles has to be determined. The circles are tangent to each other.

It is seen that if the radius of any of circles touching the point of intersection between two circles is extended it reaches the center of the other circle. Joining the centers of the three circles gives an equilateral triangle with sides equal to 4. The area that has to be determined is equal to the sum of the area of the segments of the three circles subtracted from the area of the triangle. The angle of the three segments is 60 degrees. As the area of a circle is `pi*r^2` = `4*pi` , the sum of the area of the three segments is `3*(60/360)*4*pi = 2*pi`

The area of the triangle is `sqrt 3/4*4^2` = `4*sqrt 3`

**The required area is **`4*sqrt 3 - 2*pi`